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3 years ago

What are the intercepts of the line 2x + 4y = 12?

2 answers:

8 0

For x-intercepts, you plug 0 for y and solve for x, and vice versa for y-intercepts.

X: 2X+4(0)= 12
2x= 12
12/2= 6 so the x intercept is 6.

Y: 2(0)+4Y=12
4y= 12
12/4= 3 so the y intercept is 3.

Your answer is B. <span>x-intercept: 6; y-intercept: 3</span>

Debora [2.8K]3 years ago

6 0

The answer is B. if you make the x zero to find the y-int it would be 3 and if you make the y zero to find the x-int it would be be x-int= 6 y-int=3

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Answer:

a) 0.7412 = 74.12% probability that all the three orders will be filled correctly.

b) 0.0009 = 0.09% probability that none of the three will be filled correctly

c) 0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d) 0.9991 = 99.91% probability that at least one of the three will be filled correctly

e) 0.0082 = 0.82% probability that only your order will be filled correctly

Step-by-step explanation:

For each order, there are only two possible outcomes. Either it is filled correctly, or it is not. Orders are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The percentage of orders filled correctly at Burger King was approximately 90.5%.

This means that $p = 0.905$

You and 2 friends:

So 3 people in total, which means that $n = 3$

a. What is the probability that all the three orders will be filled correctly?

This is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{3,3}.(0.905)^{3}.(0.095)^{0} = 0.7412$

0.7412 = 74.12% probability that all the three orders will be filled correctly.

b. What is the probability that none of the three will be filled correctly?

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.905)^{0}.(0.095)^{3} = 0.0009$

0.0009 = 0.09% probability that none of the three will be filled correctly.

c. What is the probability that one of the three will be filled correctly?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{3,1}.(0.905)^{1}.(0.095)^{2} = 0.0245$

0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d. What is the probability that at least one of the three will be filled correctly?

This is

$P(X \geq 1) = 1 - P(X = 0)$

With what we found in b:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0009 = 0.9991$

0.9991 = 99.91% probability that at least one of the three will be filled correctly.

e. What is the probability that only your order will be filled correctly?

Yours correctly with 90.5% probability, the other 2 wrong, each with 9.5% probability. So

$p = 0.905*0.095*0.095 = 0.0082$

0.0082 = 0.82% probability that only your order will be filled correctly

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Answer:

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Step-by-step explanation:

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